MA280 Probability
1
Overview
Probability theory:
a tool of fundamental importance to nearly all scientists,
engineers, medical practitioners jurists, and industrialists
This course provides an elementary introduction to the
theory of probability
2
The basic principle of counting
Situation:
Two experiments to be performed
Experiment 1: can result in any of m possible outcomes
For each outcome of experiment 1, experiment 2 can result
in any of the n possible outcomes.
Conclusion:
The total number of possible outcomes of the two
experiments together: mn
Proof. List all the possible outcomes of the 2 experiments.
3
Example 2a
A small community consists of 10 women, each has 3 children.
If one woman and one of her children are to be chosen as
mother and child of the year, how many different choices are
possible?
4
The generalized basic principle of counting
Situation
r experiments are to be performed.
Experiment 1 may result in n1 possible outcomes
For each of the possible outcomes for Experiment 1,
Experiment 2 may result in n2 possible outcomes
For each of the possible outcomes for Experiments 1 and
2, Experiment 3 may result in n3 possible outcomes
. . .
Conclusion
The total number of possible outcomes for the r
experiments together: n1 · n2 · · · nr .
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Example 2b
A college planning committee consists of 3 freshmen, 4
sophomores, 5 juniors, and 2 seniors. A subcommittee of 4,
consisting of 1 person from each class, is to be chosen. How
many different subcommittees are possible?
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Notation: !
Suppose there are n objects.
The total number of different permutations (ordered
arrangements) of the n objects:
n! = n(n − 1)(n − 2) · · · 3 · 2 · 1
Notes:
0! := 1
n! reads as “n factorial”
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Example 3a
How many different batting orders are possible for a baseball
team consisting of 9 players?
Solution. 9! = 9 ∗ 8 ∗ · · · ∗ 1 = 362,880.
8
Permutations when certain of the objects are
indistinguishable from one another
Example 3d How many different letter arrangements can be
formed from the letters pepper?
Solution.6!/(3!2!)
There are 6! permutations of the letters when the 3p’s and the
2e’s are distinguished from one another.
However, consider any one of these permutations for instance,
if we now permute the p’s among themselves and the e’s
among themselves,
then the resultant 3!2! arrangements would still be of the same
form when the same letters are indistinguishable from one
another
Activity: Explain the above idea to each other.
9
Permutations when certain of the objects are
indistinguishable from one another
The total number of different permutations of n objects, of
which n1 are alike, n2 are alike, · · · , nr are alike is:
n!
n1!n2! · · · nr !
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Exercise
A chess tournament has 10 competitors, of which 4 are
Russian, 3 are from the United States, 2 are from Great Britain,
and 1 is from Brazil. If the tournament result lists just the
nationalities of the players in the order in which they placed,
how many outcomes are possible?
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Combinations
Question of interest:
There are n (different) objects in total
A group is formed by selecting r objects
Determine the number of different groups that could be
formed.
Example How many different groups of 3 objects could be
selected from the 5 items, A, B, C, D and E?
Solution. 5 ∗ 4 ∗ 3/(3 ∗ 2 ∗ 1)
There are 5 ways to select the initial item, 4 ways to then select
the next item, 3 ways to select the final item.
There are thus 5 ∗ 4 ∗ 3 ways of selecting the group of 3 when
the order in which the items are selected is relevant.
However, since every group of 3–say, the group consisting of A,
B and C will be counted 6 times ( ABC, ACB, BAC, BCA, CAB,
CBA), so the group ABC (when order of selection is irrelevant)
is counted 6 times
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Forming groups of r items from a set of n objects
The number of different ways for forming a group of r items
from a set of n objects:(
n
r
)
:=
n!
(n − r)!r !
Note:
(n
r
)
reads as “n choose r "
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Example 4a
A committee of 3 is to be formed from a group of 20 people.
How many different committees are possible?
Solution.
(20
3
)
= 20!/(17!3!) = 20 ∗19 ∗18/(3 ∗2 ∗1) = 1140.
Exercise A committee of 4 is to be formed from a group of 10
people. How many different committees are possible?
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Similarly,
The number of possible ways to divide n distinct objects
into r distinct groups of respective sizes n1, n2, · · · , nr is(
n
n1,n2, · · · ,nr
)
:=
n!
n1!n2! · · · nr !
Activity: Orally design a question that requires the use of the
above formulae and ask the student next to you to answer it.
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The binomial theorem
(x + y)n =
n∑
k=0
(
n
k
)
xkyn−k
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The multinomial theorem
(x1 + x1 + · · ·+ xr )n
=
∑
(n1,··· ,nr ):n1+···+nr=n
(
n
n1,n2, · · · ,nr
)
xn11 x
n2
2 · · · xnrr
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Sample space
Sample space S: the set of all possible outcomes of an
experiment
Example 1: An experiment consists of flipping two coins.
The sample space
S = {(h,h), (h, t), (t ,h), (t , t)},
where t = tail and h = head .
Example 2: An experiment consists of tossing two dice.
The sample space:
S = {(i , j) : i , j = 1,2, · · · ,6}
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Event
Event: any subset of the sample space
An event is a set consisting of some possible outcomes of
the experiment.
If the outcome of the experiment is contained in then we
say that has occurred.
Example 1: An experiment consists of flipping two coins.
The sample space
S = {(h,h), (h, t), (t ,h), (t , t)},
E = {(h,h), (h, t)} is an event (the event that a head
appears on the first coin).
Example 2: An experiment consists of tossing two dice.
The sample space:
S = {(i , j) : i , j = 1,2, · · · ,6}
E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} is an event
(the event that the sum of the dice equals 7)
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Union
Union of events E and F : E ∪ F
E ∪ F : the event that consists of all outcomes that are
either in E
or in F
or in both E and F
Venn diagram
Example 1: Suppose E = {(h,h), (h, t)} and
F = {(t ,h), (h,h)}.
Then E ∪ F = {(h,h), (t ,h), (h, t)}.
The union, ∪∞n=1En: the event that consists of all outcomes
that are in at least one of the sets, E1, · · · , En.
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Intersection
Intersection of events E and F : E ∩ F or EF
E ∩ F : the event that consists of all outcomes that are
both in E and in F
Venn diagram
Example 1: Suppose E = {(h,h), (h, t)} and
F = {(t ,h), (h,h)}.
Then E ∩ F = {(h,h)}.
The intersection, ∩∞n=1En: the event that consists of those
outcomes that are in all of the events, E1, · · · , En.
If E ∩ F does not contain any outcomes, it is the null event
and is denoted by ∅
If E ∩ F = ∅, then E and F are said to be mutually
exclusive
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Complementation
Complementation. Ac , the complement of A, is the set of
all elements that are not in A:
Ac = {x ; x ∈ A}
Venn diagram
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Exercise
Exercise: Suppose A = {(h,h), (h, t)}, B = {(t ,h)} and
C = {(h,h), (h, t), (t , t)}. Find A ∩ B, A ∪ C, A ∩ C and Cc .
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Some rules for the operations of events
Commutative laws
E ∪ F = F ∪ E Exercise Show this is by using Venn diagrams
E ∩ F = F ∩ E Exercise Show this is by using Venn diagrams
Associative laws
E ∪ F ∪G = E ∪ (F ∪G)
E ∩ F ∩G = E ∩ (F ∩G)
Distributive laws:
(E ∪F )∩G = (E ∩G)∪(F ∩G) Exercise Show this is by using Venn diagrams
(E ∩F )∪G = (E ∪G)∩(F ∪G) Exercise Show this is by using Venn diagrams
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DeMorgan’s laws
DeMorgan’s laws:
a) (∪ni=1Ei)c = ∩ni=1Eci
b) (∩ni=1Ei)c = ∪ni=1Eci
Proof to a):
Suppose x ∈ (∪ni=1Ei)c
=⇒ x /∈ ∪ni=1Ei
=⇒ x /∈ Ei for all i = 1,2, · · ·
=⇒ x ∈ Eci for all i = 1,2, · · ·
=⇒ x ∈ ∩ni=1Eci for all i = 1,2, · · ·
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Axioms of Probability
P(E): the probability of the event E
Each probability measure P should satisfies the following
axioms
Axiom 1: 0 ≤ P(E) ≤ 1
Axiom 2: P(S) = 1 where S is the sample space
Axiom 3: For any sequence of mutually exclusive events,
E1, E2, . . . (EiEj = ∅ when i ̸= j),
P
(∞⋃
i=1
Ei
)
=
∞∑
i=1
P(Ei)
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Example 3b
Example 3b
Suppose a die (with 6 sides) is rolled and that all 6 sides are
equally likely to appear. What is the probability of rolling an
even number?
Answer.
The event of rolling even number: {2,4,6}
We have
P({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1
6
.
Note {2}, {4} and {6} is a sequence of mutually exclusive
events.
So by Axiom 3, the prob. of rolling an even number:
P({2,4,6}) = P({2}∪{4}∪{6}) = P({2})+P({4})+P({6}) = 1
2
.
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Some simple propositions
Proposition 4.1: P(Ec) = 1− P(E), or equivalently,
P(E) + P(Ec) = 1.
Proposition 4.2 If E ⊂ F , then P(E) ≤ P(F ).
Proof.
Note F = (E ∪ Ec)F = E ∪ (EcF ) (Exercise Show this in
Venn diagrams.)
Note E and EcF are mutually exclusive
By Axiom 3:
P(F ) = P(E) + P(EcF ) ≥ P(E)
(since P(EcF ) ≥ 0 by Axiom 1)
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Some simple propositions
Proposition 4.3: P(E ∪ F ) = P(E) + P(F )− P(EF ).
Proof.
E ∪ F = S(E ∪ F ) = (E ∪ Ec)(E ∪ F ) = E ∪ (EcF )
(Exercise Show the above in Venn diagrams.)
E and EcF mutually exclusive (by Axiom 3) =⇒
P(E ∪ F ) = P(E ∪ (EcF )) = P(E) + P(EcF )
Note F = (EF ) ∪ (EcF ) (Exercise Show this in Venn
diagrams.)
(by Axiom 3) =⇒
P(F ) = P(EF ) + P(EcF ), i.e., P(EcF ) = P(F )− P(EF )
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